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Program to find longest even value path of a binary tree in Python
Suppose we have a binary tree, we have to find the longest path consisting of even values between any two nodes in the tree.
So, if the input is like
then the output will be 5 as the longest path is [10, 2, 4, 8, 6].
To solve this, we will follow these steps −
ans := 0
Define a function find() . This will take node
if node is null, then
return (-1, -1)
leftCnt := maximum of returned value of find(left of node) + 1
rightCnt := maximum of returned value of find(right of node) + 1
if value of node is even, then
ans := maximum of ans and (leftCnt + rightCnt + 1)
return (leftCnt, rightCnt)
otherwise,
ans := maximum of ans, leftCnt and rightCnt
return (-1, -1)
From the main method do the following −
find(root)
return ans
Let us see the following implementation to get better understanding −
Example
class TreeNode: def __init__(self, val, left=None, right=None): self.val = val self.left = left self.right = right class Solution: def solve(self, root): ans = 0 def find(node): nonlocal ans if not node: return -1, -1 leftCnt = max(find(node.left)) + 1 rightCnt = max(find(node.right)) + 1 if node.val % 2 == 0: ans = max(ans, leftCnt + rightCnt + 1) return leftCnt, rightCnt else: ans = max(ans, max(leftCnt, rightCnt)) return -1, -1 find(root) return ans ob = Solution() root = TreeNode(2) root.left = TreeNode(10) root.right = TreeNode(4) root.right.left = TreeNode(8) root.right.right = TreeNode(2) root.right.left.left = TreeNode(6) print(ob.solve(root))
Input
root = TreeNode(2) root.left = TreeNode(10) root.right = TreeNode(4) root.right.left = TreeNode(8) root.right.right = TreeNode(2) root.right.left.left = TreeNode(6)
Output
5
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